题目 Problem (a) Show that 3tan2θ+5sin2θ=8sin2θ−5sin4θ1−sin2θ.3\tan^2\theta + 5\sin^2\theta = \frac{8\sin^2\theta - 5\sin^4\theta}{1 - \sin^2\theta}.3tan2θ+5sin2θ=1−sin2θ8sin2θ−5sin4θ. [3] (b) Hence solve the equation 3tan2θ+5sin2θ=93\tan^2\theta + 5\sin^2\theta = 93tan2θ+5sin2θ=9 for 0∘<θ<270∘0^\circ < \theta < 270^\circ0∘<θ<270∘. [4] 题目中文翻译 (a) 证明 3tan2θ+5sin2θ=8sin2θ−5sin4θ1−sin2θ3\tan^2\theta + 5\sin^2\theta = \frac{8\sin^2\theta - 5\sin^4\theta}{1 - \sin^2\theta}3tan2θ+5sin2θ=1−sin2θ8sin2θ−5sin4θ [3] (b) 由此解方程 3tan2θ+5sin2θ=93\tan^2\theta + 5\sin^2\theta = 93tan2θ+5sin2θ=9，其中 0∘<θ<270∘0^\circ < \theta < 270^\circ0∘<θ<270∘。 [4] 解答
